Question: Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = -\dfrac{2}{3}x + 4}\enspace$ and passes through the point ${(-3, -3)}$. {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9}
Answer: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${-\dfrac{2}{3}}$ , and its negative reciprocal is ${\dfrac{3}{2}}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = \dfrac{3}{2}x + b}\enspace$ We can plug our point, $(-3, -3)$ , into this equation to solve for ${b}$ , the y-intercept. $-3 = {\dfrac{3}{2}}(-3) + {b}$ $-3 = -\dfrac{9}{2} + {b}$ $-3 + \dfrac{9}{2} = {b} = \dfrac{3}{2}$ The equation of the perpendicular line is $\enspace {y = \dfrac{3}{2}x + \dfrac{3}{2}}\enspace$. ${m = \dfrac{3}{2}, \enspace b = \dfrac{3}{2}}$